結果
問題 | No.270 next_permutation (1) |
ユーザー |
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提出日時 | 2015-08-22 00:05:45 |
言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 77 ms / 2,000 ms |
コード長 | 2,271 bytes |
コンパイル時間 | 1,707 ms |
コンパイル使用メモリ | 163,140 KB |
実行使用メモリ | 6,820 KB |
最終ジャッジ日時 | 2024-12-18 02:07:56 |
合計ジャッジ時間 | 2,939 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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ファイルパターン | 結果 |
---|---|
other | AC * 15 |
ソースコード
#include <bits/stdc++.h>#define rep(i, a) rep2 (i, 0, a)#define rep2(i, a, b) for (int i = (a); i < (b); i++)#define repr(i, a) repr2 (i, 0, a)#define repr2(i, a, b) for (int i = (b) - 1; i >= (a); i--)#define asn(a, b, c) fill_n(&(b), sizeof(a) / sizeof(b), c)using namespace std;typedef long long ll;const ll inf = 1e9;const ll mod = 1e9 + 7;template <class BidirectionalIterator, class Compare>int next_permutation2(BidirectionalIterator first, BidirectionalIterator last, Compare comp){if (first == last)return false;BidirectionalIterator i = first;++i;if (i == last)return 0;i = last;--i;for(;;) {BidirectionalIterator ii = i;--i;if (comp(*i, *ii)) {BidirectionalIterator j = last;while (!comp(*i, *--j)) {}std::swap(*i, *j);std::reverse(ii, last);return i - first;}if (i == first) {std::reverse(first, last);return i - first;}}}template <class BidirectionalIterator>int next_permutation2(BidirectionalIterator first, BidirectionalIterator last){typedeftypename std::iterator_traits<BidirectionalIterator>::value_typevalue_type;return next_permutation2(first, last, std::less<value_type>());}int N, K;int p[100000], B[100000];struct BIT {int size;vector<ll> bit;BIT(int size) : size(size), bit(size + 1) {}void add(int k, ll x) {while (k <= size) {bit[k] += x;k += k & -k;}}ll sum(int k) {ll res = 0;while (k > 0) {res += bit[k];k -= k & -k;}return res;}void update(int k, ll x) {ll g = sum(k) - sum(k - 1);add(k, x - g);}};int main() {cin >> N >> K;rep (i, N) cin >> p[i];rep (i, N) cin >> B[i];BIT bit(N * 2);ll ans = 0;if (K == 0) {cout << 0 << endl;return 0;}rep (i, N) {bit.add(i + 1, abs(B[i] - p[i]));}ans += bit.sum(N + 1);rep (ii, K - 1) {int k = next_permutation2(p, p + N);rep2 (i, k, N) {bit.update(i + 1, abs(B[i] - p[i]));}ans += bit.sum(N + 1);}cout << ans << endl;return 0;}